Timeline for Bialgebra pairing on ring polynomial $K[x]$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| May 5, 2017 at 23:08 | comment | added | Vladimir Dotsenko | @UmarMuhammadFaruq If the ground field is of characteristic $p$, then the kernel of the pairing is the ideal generated by $x^p$, since $n!$ is divisible by $p$ for all $n\ge p$. | |
| May 5, 2017 at 22:59 | comment | added | unknown | Good. If you don't mind I will also like to know what if $char k = p>0$ because It seems from the compuatation above that $ker(t)=x^p$, right? thanks. | |
| May 5, 2017 at 22:50 | history | edited | Vladimir Dotsenko | CC BY-SA 3.0 |
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| May 5, 2017 at 22:49 | comment | added | Vladimir Dotsenko | @UmarMuhammadFaruq this calculation does indeed show that the pairing is nondegenerate in char 0. It assumes the "standard" bialgebra structure on $k[x]$ for which $x$ is a primitive element, and nothing else. | |
| May 5, 2017 at 22:32 | comment | added | unknown | Thanks. That's really helpful. Also can I use the primitive properties to say that $t$ is non-degenerate if char $k=0$?. | |
| May 5, 2017 at 22:04 | comment | added | Vladimir Dotsenko | @მამუკაჯიბლაძე : I assume that the author of the question means the standard bialgebra structure on the polynomial ring when writing "the polynomial ring". I still think that "the primitive property is not needed here" means that the original version of the question was overloaded by unnecessary notation, not that $x$ is not assumed primitive. | |
| May 5, 2017 at 21:34 | comment | added | მამუკა ჯიბლაძე | The answer is perfect for the previous version of the question. Note that now it is not presumed anymore that $x$ is primitive. | |
| May 5, 2017 at 21:10 | history | answered | Vladimir Dotsenko | CC BY-SA 3.0 |